Convergence of $f_n(x)=n^2x(1-x^2)^n$ on $[0,1)$

Question

First I note that $limsup_{x\in[0,1)}=n^2$, although I do not know how to prove this rigorously. This leads me to believe that the function $f_n(x)=n^2x(1-x^2)^n$ does not uniformly convergence on $[0,1)$, because this implies that given a fixed n, $\forall \epsilon>0 , \exists x\in[0,1)$ s.t $|f_n(x)-n^2|<\epsilon$, and so this series can't uniformly converge to a certain value because $n^2$ is different (monotonely increasing) for each n.

If it's not obvious, I need a bit of help clarifying things.

Answer 1

Your sequence $(f_n)_n$ converges to $0$ pointwise. Namely, if $x \in (0,1)$ then $1-x^2 \in (0,1)$ so $\lim_{n\to\infty} n^2 x(1-x^2)^n = 0$ because the exponential function decays to zero faster than the growth of any polynomial. On the other hand, if $x = 0$ then $\lim_{n\to\infty} n^2 x(1-x^2)^n = \lim_{n\to\infty} 0 = 0$.

To find it the convergence is uniform, we need to calculate $$\sup_{x\in[0,1)} \left|f_n(x)\right| = \sup_{x \in [0,1)} n^2x(1-x^2)^n$$

To find the maximum, we calculate the derivative

$$0 =f'(x) = n^2\left[(1-x^2)^n + nx(1-x^2)^{n-1}\cdot(-2x)\right]$$

giving $$(1-x^2)^n = 2nx^2(1-x^2)^{n-1}\implies x = \frac{1}{\sqrt{2n+1}}$$

So $$\sup_{x \in [0,1)} n^2x(1-x^2)^n = \max\left\{f_n(0), f_n\left(\frac{1}{\sqrt{2n+1}}\right)\right\} = \frac{n^2}{\sqrt{2n+1}} \left(1-\frac{1}{1 + 2 n}\right)^{n}$$

which clearly diverges to $+\infty$ as $n\to\infty$:

\begin{align} \lim_{n\to\infty}\frac{n^2}{\sqrt{2n+1}} \left(1-\frac{1}{2n+1}\right)^{n} &= \lim_{n\to\infty} \frac{n^2}{\sqrt{2n+1}}\left[\left(1-\frac1{2n+1}\right)^{2n+1}\right]^{\frac{n}{2n+1}} = (+\infty)\cdot e^{-1/2} = +\infty \end{align}

Therefore, the convergence $f_n \to 0$ is not uniform.

Answer 2

Either you already have or should be able to prove that $f_n(x) \to 0$ pointwise.

Convergence is clearly not uniform since

$$\sup_{x \in [0,1)} f_n(x) \geqslant f_n(1/\sqrt{n}) = n^{3/2} (1 - 1/n)^n \to \infty$$

Note that $(1-1/n)^n \to e^{-1}$.